Integrable function whose Fourier transform is not integrable I am looking for an example of a function $f: \mathbb R \rightarrow \mathbb R$ such that $f \in L^1$ in the sense that $\int_{\mathbb R} |f| < \infty$ but its Fourier transform $\hat f$ is not in $L^1$. Does anyone have one?
Thanks.
 A: Added: The function $f(x) = \vert x \vert^{-1/2} \mathrm{e}^{-\vert x \vert}$ is a simple example (much simpler than the original example I proposed). Its Fourier transform is:
$$
   \hat{f}(\omega) = \sqrt{\frac{1}{\sqrt{\omega ^2+1}}+\frac{1}{\omega ^2+1}}
$$
and has asymptote $\hat{f}(\omega) \sim \vert \omega \vert^{-1/2}$ for large $\vert \omega \vert$, thus $\hat{f} \not\in L^1$.
 Original example:

An example would be 
$$
   f(x) = \left\{ \begin{array}{cc} 
     x^{-1/4} \mathrm{e}^{-x} & x > 0 \\
     \vert x \vert^{-1/2} \mathrm{e}^{x} & x < 0
    \end{array} \right.
$$
It is clear that $\int_\mathbb{R} \vert f(x) \vert \mathrm{d} x < \infty$. The Fourier transform
$$
   \hat{f}(\omega) = 
\frac{\sqrt{1-i \omega }-\sqrt{1+i \omega }}{ \sqrt{8 (1+\omega
   ^2)}}+\frac{1}{2} \sqrt{\frac{1}{\sqrt{\omega ^2+1}}+\frac{1}{\omega
   ^2+1}}+\frac{\Gamma \left(\frac{3}{4}\right)}{\sqrt{2 \pi } \, (1-i \omega )^{3/4}}
$$
The integral $\int_\mathbb{R} \vert \hat{f}(\omega) \vert \mathrm{d} \omega $ diverges because $\hat{f}(\omega) \sim \vert \omega \vert^{-\frac{1}{2}}$.
A: Note that any function whose Fourier transform is in $L^1$ must be equal to a continuous function almost everywhere, since $\mathcal F^{-1}(\mathcal F(f)) = f$ a.e. in this case. This follows from the inversion formula and because the Fourier transform of a function is continuous.
This gives us many examples of functions you are looking for. For example, $f(x) = \chi_{[-1,1]}(x)$ must necessarily be such a function.
